The following article shows how given an integer list, we can produce the sum of all its odd and even digits.
Input : test_list = [345, 893, 1948, 34, 2346] Output : Odd digit sum : 36 Even digit sum : 40 Explanation : 3 + 5 + 9 + 3 + 1 + 9 + 3 + 3 = 36, odd summation.
Input : test_list = [345, 893] Output : Odd digit sum : 20 Even digit sum : 12 Explanation : 4 + 8 = 12, even summation.
Method 1 : Using loop, str() and int()
In this, we first convert each element to string and then iterate for each of its element, and add to respective summation by conversion to integer.
Python3
# initializing listtest_list = [345, 893, 1948, 34, 2346]# printing original listprint("The original list is : " + str(test_list))odd_sum = 0even_sum = 0for sub in test_list: for ele in str(sub): # adding in particular summation according to value if int(ele) % 2 == 0: even_sum += int(ele) else: odd_sum += int(ele)# printing resultprint("Odd digit sum : " + str(odd_sum))print("Even digit sum : " + str(even_sum)) |
The original list is : [345, 893, 1948, 34, 2346] Odd digit sum : 36 Even digit sum : 40
Time Complexity: O(n*n) where n is the length of the input list test_list, and m is the maximum number of digits in any element of the list.
Auxiliary Space:The auxiliary space complexity of this program is O(1), because the program uses a constant amount of extra space to store the odd_sum and even_sum variables, regardless of the size of the input list.
Method 2: Using loop and sum()
In this, we perform task of getting summation using sum(), and loop is used to perform the task of iterating through each element.
Python3
# initializing listtest_list = [345, 893, 1948, 34, 2346]# printing original listprint("The original list is : " + str(test_list))odd_sum = 0even_sum = 0for sub in test_list: # sum() used to get summation of even and odd elements odd_sum += sum([int(ele) for ele in str(sub) if int(ele) % 2 == 1]) even_sum += sum([int(ele) for ele in str(sub) if int(ele) % 2 == 0])# printing resultprint("Odd digit sum : " + str(odd_sum))print("Even digit sum : " + str(even_sum)) |
The original list is : [345, 893, 1948, 34, 2346] Odd digit sum : 36 Even digit sum : 40
Time complexity: O(N * M).
Auxiliary space: O(1).
Method 3: Using list comprehension
Python3
test_list = [345, 893, 1948, 34, 2346]odd_sum = 0even_sum = 0odd_sum += sum([int(ele) for sub in test_list for ele in str(sub) if int(ele) % 2 == 1])even_sum += sum([int(ele) for sub in test_list for ele in str(sub) if int(ele) % 2 == 0])print("Odd digit sum : " + str(odd_sum))print("Even digit sum : " + str(even_sum)) |
Odd digit sum : 36 Even digit sum : 40
Time complexity: O(nm).
Auxiliary space: O(1).
Method 4: Using the enumerate function
Python3
test_list = [345, 893, 1948, 34, 2346]odd_sum = 0even_sum = 0odd_sum += sum([int(ele) for i, sub in enumerate(test_list) for ele in str(sub) if int(ele) % 2 == 1])even_sum += sum([int(ele) for i, sub in enumerate(test_list) for ele in str(sub) if int(ele) % 2 == 0])print("Odd digit sum : " + str(odd_sum))print("Even digit sum : " + str(even_sum)) |
Odd digit sum : 36 Even digit sum : 40
The time complexity of the given program is O(n*k), where n is the number of elements in the list and k is the maximum number of digits in any element of the list.
The space complexity of the given program is O(1) because it uses a constant amount of extra space regardless of the size of the input.
Method 5: Using recursion
Python3
def digit_sum(sub, even_sum, odd_sum): if not sub: return (even_sum, odd_sum) else: ele = sub.pop() if int(ele) % 2 == 0: even_sum += int(ele) else: odd_sum += int(ele) return digit_sum(sub, even_sum, odd_sum)test_list = [345, 893, 1948, 34, 2346]# printing original listprint("The original list is : " + str(test_list))even_sum = 0odd_sum = 0for sub in test_list: sub = list(str(sub)) even_sum, odd_sum = digit_sum(sub, even_sum, odd_sum)# printing resultprint("Odd digit sum : " + str(odd_sum))print("Even digit sum : " + str(even_sum))#This article is published by Vinay Pinjala. |
The original list is : [345, 893, 1948, 34, 2346] Odd digit sum : 36 Even digit sum : 40
Time Complexity: O(n)
Auxiliary Space: O(n)
Approach: Iterative Digit Summing Approach
Steps:
- Initialize two variables, even_sum and odd_sum, to 0.
- Iterate over each number in the list using a while loop until the number becomes 0.
- Extract the last digit of the number using modulo operator and add it to even_sum if it is even, else add it to odd_sum.
- Remove the last digit from the number using integer division operator.
- Return both even_sum and odd_sum.
Python3
def sum_of_even_odd_digits(test_list): even_sum = 0 odd_sum = 0 for num in test_list: while num != 0: digit = num % 10 if digit % 2 == 0: even_sum += digit else: odd_sum += digit num //= 10 return even_sum, odd_sumtest_list = [345, 893, 1948, 34, 2346]even_sum, odd_sum = sum_of_even_odd_digits(test_list)print("Odd digit sum :", odd_sum)print("Even digit sum :", even_sum) |
Odd digit sum : 36 Even digit sum : 40
Time Complexity: O(n * d), where n is the number of elements in the list and d is the maximum number of digits in any number in the list.
Auxiliary Space: O(1)
